The extendability of matchings in strongly regular graphs

A graph G of even order v is called t-extendable if it contains a perfect matching, t < v/2 and any matching of t edges is contained in some perfect matching. The extendability of G is the maximum t such that G is t-extendable. In this paper, we study the extendability properties of strongly regular graphs. We improve previous results and classify all strongly regular graphs that are not 3-extendable. We also show that strongly regular graphs of valency k > 3 with λ > 1 are bk/3c-extendable (when μ 6 k/2) and d 4 e-extendable (when μ > k/2), where λ is the number of common neighbors of any two adjacent vertices and μ is the number of common neighbors of any two non-adjacent vertices. Our results are close to being best possible as there are strongly regular graphs of valency k that are not dk/2e-extendable. We show that the extendability of many strongly regular graphs of valency k is at least dk/2e − 1 and we conjecture that this is true for all primitive strongly regular graphs. We obtain similar results for strongly regular graphs of odd order.


Introduction
A set of edges M of a graph G is a matching if no two edges of M share a vertex. A matching M is perfect if every vertex is incident with exactly one edge of M . A matching is near perfect if all but one of the vertices of G are incident with edges of the matching. A graph G of even order v is called t-extendable if it contains at least one perfect matching, and Lou [1]. Many strongly regular graphs have trivial automorphism groups and our techniques are different than the ones used for vertex-transitive graphs.
In this paper, we show that every connected (v, k, λ, µ)-srg of valency k 5 is 3extendable with exception of the complete 4-partite graph K 2,2,2,2 (the (8,6,4,6)-srg), the complement of the Petersen graph (the (10, 6, 3, 4)-srg) and the Shrikhande graph (one of the two (16, 6, 2, 2)-srgs). We also prove that any connected (v, k, λ, µ)-srg with λ 1 is k/3 -extendable when µ k/2 and k+1 4 -extendable when µ > k/2. This result is close to being best possible as we will prove that many connected strongly regular graphs with valency k, λ 1 are not k/2 -extendable. On the other hand, we also determine the extendability of many families of strongly regular graphs including Latin square graphs, block graphs of Steiner systems, triangular graphs, lattice graphs and all known triangle-free strongly regular graphs. For each graph of valency k that we considered, the extendability is at least k/2 − 1. We conjecture that this is true for all primitive strongly regular graphs. We also obtain similar results for strongly regular graphs of odd order.

Main tools
In this section, we introduce the notation used in our paper and describe the main tools used in our proofs. For undefined terms, see Brouwer and Haemers [7]. Let o(G) denote the number of components of odd order of a graph G. If S is a subset of vertices of G, then G−S denotes the subgraph of G obtained by deleting the vertices in S. The independence number of G will be denoted by α(G). If S and T are vertex disjoint subsets of a graph, let e(S, T ) denote the number of edges with one endpoint in S and the other in T . Let N (T ) denote the set of vertices outside T that are adjacent to at least one vertex of T . When T = {x}, let N (x) = N ({x}). If x is a vertex of a strongly regular graph G, let N 2 (x) = V (G) \ ({x} ∪ N (x)); the first subconstituent Γ 1 (x) of x is the subgraph of G induced by N (x) and the second subconstituent Γ 2 (x) of x is the subgraph of G induced by N 2 (x). For a (v, k, λ, µ)-srg G, let k = θ 1 θ 2 . . . θ v denote the eigenvalues of its adjacency matrix. It is known that G has exactly three distinct eigenvalues k, θ 2 and θ v with θ 2 + θ v = λ − µ and θ 2 θ v = µ − k (see [7,24] for example). The complete multipartite graph which is the complement of m disjoint copies of K a will be denoted by K a×m .
Lemma 4 (Yu [39]). Let t 1 be an integer and G be a factor-critical graph. The graph G is not t-near-extendable if and only if it contains a subset of vertices S such that S contains t independent edges, |S| 2t + 1, and o(G − S) |S| − 2t + 1.
Theorem 5 (Brouwer and Mesner [10]). If G is a primitive strongly regular graph of valency k, then G is k-connected. Any disconnecting set of size k must be the neighborhood of some vertex. Lemma 6. If G is a distance-regular graph of degree k 2 and diameter D 3, then for any x ∈ V (G), the subgraph induced by the vertices at distance 2 or more from x, is connected.
Proof. As D 3, G contains an induced path P 4 with 4 vertices. By eigenvalue interlacing, the second largest eigenvalue θ 2 (G) of G, is at least θ 2 (P 4 ) = −1+ √ 5 2 > 0. Cioabȃ and Koolen [17] proved that if the i-th entry of the standard sequence corresponding to the second largest eigenvalue of a distance-regular graph G is positive, then for any vertex x, the subgraph of G induced by the vertices at distance at least i from x is connected. The second entry of the standard sequence corresponding to θ 2 (G) is θ 2 (G)/k > 0 and this finishes our proof.
Lemma 8. Let G be a (v, k, λ, µ)-srg. If T is an independent set, then . Counting the edges between T and N (T ), we have |T |k = |N (T )|d. Counting the 3-subsets of the form {x, y, z} such that x, y ∈ T, z ∈ N (T ), x ∼ z, y ∼ z, we get that Combining these equations, we obtain that (|T | − 1)µ k k|T | |N (T )| − 1 which implies the desired inequality |N (T )| the electronic journal of combinatorics 21(2) (2014), #P2.34 Note that the result of Lemma 8 is better than the one obtained by applying Bonfer- Lemma 9. Let G be a primitive (v, k, λ, µ)-srg with λ 1. If T is an independent set, then Proof then the previous arguments and Lemma 8 imply that Note that if λ 1 and µ k/2, then Lemma 9 implies that for any independent set of vertices T .
Lemma 10 (Lemma 2.3 [16]). Let G be a connected (v, k, λ, µ)-srg. Let S be a disconnecting set of vertices of G and denote by A a subset of vertices of G that induces a connected subgraph of G − S and B := V (G) \ (A ∪ S). If |A| = a and |B| = b, then The following lemma extends Theorem 5.1 of [18].
Lemma 11. Let G be a primitive (v, k, λ, µ)-srg. If A is a subset of vertices with 3 |A| v/2 and A c denotes its complement, then Proof. If k = 3, then G is K 3,3 or the Petersen graph. If G is K 3,3 , the proof is immediate. If G is the Petersen graph, and A is a subset of vertices with 3 |A| 4, then the number of edges contained in A is at most |A| − 1 and therefore e(A, A c ) 3|A| − 2(|A| − 1) = |A| + 2 5. If |A| = 5, then the number of edges inside A is at most |A| = 5 and therefore e(A, A c ) 3|A| − 2|A| = |A| = 5. If k = 4, then G is K 4,4 , K 2,2,2 or the Lattice graph L 2 (3) which is the unique (9, 4, 1, 2)-SRG. If G is K 4,4 or K 2,2,2 , the proof is immediate. If the electronic journal of combinatorics 21(2) (2014), #P2.34 G is the Lattice graph L 2 (3), and A is a subset of vertices with |A| = 3, then e(A, A c ) 6 with equality if and only if A induces a clique of order 3. If |A| = 4, then the number of edges contained in A is at most 4 and therefore, e(A, A c ) 4|A| − 8 = 8.
Assume k 5. If |A| k − 2, then e(A, A c ) |A|(k − |A| + 1) 3(k − 2). Assume |A| = k−1. If every vertex of A has at least 3 neighbors outside A, then e(A, A c ) 3(k−1). Otherwise, there exists a vertex x ∈ A that has exactly 2 neighbors outside A. Therefore, (see [7,Corollary 4.8.4] or [30]). If G is a conference graph of parameters (4t + 1, 2t, t − 1, t), then k − θ 2 = 4t+1− √ 4t+1 2 > 6 for t 4 and consequently, e(A, A c ) > 3k. If t = 3, G has parameters (13,6,2,3) and therefore, e(A, A c ) If G is not a conference graph and k − θ 2 6, then e(A, A c ) 3k and we are done again. The only case left is when G is not a conference graph and k − θ 2 5. In this case, the eigenvalues of G are integers, θ 2 k − 5 and θ v −2 as G is not a complete graph.

Lemma 12.
If G is a (v, k, λ, µ)-srg of even order with independence number 2, then the extendability of G is k 2 − 1. Proof. The fact that the extendability is at least k 2 − 1 follows from Lemma 3. If G is imprimitive, then G must be K 2×m for some m and the conclusion will follow from Section 3.1. Assume that G is primitive and α(G) = 2. For any vertex x ∈ V , let ). The second subconstituent Γ 2 (x) must be a complete graph with k − µ + 1 vertices. As the clique number is at most λ + 2, we have θ 2 (−θ v ) = k − µ λ + 1. Since θ v −2, we obtain that θ 2 λ+1 2 λ − 1 (when λ 3). The first subconstituent Γ 1 (x) is λ-regular with second largest eigenvalue at most θ 2 . By [15], Γ 1 (x) contains a matching of size k/2 . If k is even, then this matching cannot be extended to a maximum matching of G. If k is odd, one can add one disjoint edge to this matching the electronic journal of combinatorics 21(2) (2014), #P2.34 such that the result matching of size k 2 cannot be extended to a maximum matching of G. It is easy to see that when λ is 1 or 2, Γ 1 (x) contains a perfect matching or an almost perfect matching.
Lemma 15. Let G be a (v, k, λ, µ)-srg with v even, λ = 0 and k 7. If A is a subset of vertices such that 5 |A| v − k − 1 and |A| is odd, then e(A, A c ) 5k − 12.
Proof. Assume that 5 |A| 2k − 5. As G is triangle-free, so is the subgraph induced by A. By Turán's Theorem, the number of edges inside A is at most . The minimum is attained at |A| = 5 or 2k − 5. In either case, we have e(A, A c ) 5k − 12.
Lemma 16. Let G be a primitive (v, k, λ, µ)-srg and S be a disconnecting set of vertices. If G − S contains at least two singleton components, then S contains at least µ(k − µ) edges.
the electronic journal of combinatorics 21(2) (2014), #P2.34 Proof. Let x and y be two singleton components of G − S. Then N (x) ∪ N (y) ⊆ S. If z ∈ N (x) \ N (y), then z and y are non adjacent and they have exactly µ common. So, z is adjacent to at least µ vertices of S, and there are |N (x) \ N (y)| = k − µ such z. By the same argument, each vertex inside N (y) \ N (x) is adjacent to at least µ vertices of S. Thus, 2e(S) 2µ(k − µ).
3 The extendability of strongly regular graphs 3.1 Imprimitive strongly regular graphs A strongly regular graph is imprimitive if it, or its complement, is disconnected. The only imprimitive strongly regular graphs are disjoint unions of cliques of the same order and their complements (complete multipartite regular graphs). A disjoint union mK a of some number m 2 of cliques K a does not contain a perfect matching nor a near perfect matching if a is odd. If a is even, the extendability of this graph is am/2 − 1. The complete multipartite graph K a×m (which is the complement of mK a ) has extendability

Lower bounds for the extendability of strongly regular graphs
In this section, we classify the primitive strongly regular graphs of even order that are not 3-extendable. We first provide results giving some general lower bounds for the extendability of a primitive strongly regular graphs.
Note that Chen [14] proved that every (2t + n − 2)-connected K 1,n -free graph of even order is t-extendable (see also Aldred and Plummer [2] for extensions of Chen's result). When λ 1, every (v, k, λ, µ)-srg is k-connected and K 1, k/2 +1 -free. If we let t = 1 2 k+2 2 and n = k/2 +1, then Chen's result implies that such strongly regular graph is 1 2 k+2 2 extendable. This is the same as our result when θ 2 = 1, µ > k/2 and λ k−3 2 . For other cases, our lower bound is better than Chen's result. Note that Chen's bound can be improved if one has a better bound than k/2 for the independence number of the first subconstituents of the strongly regular graph.
Proof. If G is not n-extendable, by Lemma 3, there is a vertex set S with s vertices such that S contains n independent edges, and G − S has at least s − 2n + 2 odd components. Let O 1 , O 2 , . . . , O r be all the odd components of G − S, with r s − 2n + 2. Let a 0 denote the number singleton components among O 1 , . . . , O r . Counting the number of edges between S and O 1 ∪ · · · ∪ O r and using Lemma 11, we get the following This inequality is equivalent to and since s − a k − 1 (see the remark following Lemma 9), we obtain that This is a contradiction with n = k 2 −k−3 the electronic journal of combinatorics 21(2) (2014), #P2.34 Corollary 20. Any primitive (v, k, λ, µ)-srg with v even, λ 1 and k 8 is 3-extendable.
Proof. We show that G is 3-extendable by contradiction. Assume that G is not 3extendable. Lemma 3 implies that G has a vertex subset S, such that S contains 3 independent edges, and o(G − S) |S| − 4. Let S be such disconnecting set with maximum size. We first claim that any non-singleton component of G − S cannot induce a bipartite graph. If that was the case, the respective component would have two partitions X and Y . Assume that |X| > |Y |, then define S = S ∪ Y . Then |S | > |S| and o(G − S ) |S | − 4, contradicting the maximality of |S|. Note that G − S cannot contain exactly 3 vertices, because G is triangle free and any component with 3 vertices must be a path, which is bipartite. By similar argument, G − S contains no even components. If it contains a even component, we can put one vertex of this even component into S, which make |S| larger and S still satisfy |S| − 4 3, G − S has at least two non-singleton components, thus |S| k + 1. By using Lemma 15 and counting the edges between S and O 1 ∪ · · · ∪ O r , we will get the following, Thus, 2k 2 − 16k + 27 0, contradiction with k 8. If G − S has at least one non-singleton components and at least two singleton components, by using Lemma 15 and Lemma 16 and counting the edges between S and O 1 ∪ · · · ∪ O r , we will get that which will yield another contradiction with k 8.
Proof. If k 8, then G is 3-extendable by Corollary 20 and Theorem 21. There are two primitive parameter sets with v even, λ 1 and 5 k 7: (10, 6, 3, 4) and (16,6,2,2). There is a unique (10, 6, 3, 4)-srg, the complement of Petersen graph or the triangular graph T (5). Theorem 24 will show that the extendability of this graph is 2. There are two non-isomorphic strongly regular graphs with parameter set (16,6,2,2). One is the Shrikhande graph (see [7, page 123] for a description) and the other is the line graph of K 4,4 . In the Shrikhande graph, the first subconstituent of a fixed vertex is isomorphic to the cycle C 6 and thus, contains a matching of size 3. This matching is not contained in any perfect matching. Thus, the Shrikhande graph is not 3-extendable; by Lou and Zhu [29], the extendability of the Shrikhande graph is 2. Proposition 30 will show that the extendability of the line graph of K 4,4 is 3. To finish the proof, the only strongly regular graph with 5 k 7 and λ = 0 is the folded 5-cube whose extendability is 3 (see Theorem 32).
By a more extensive case analysis which we omit here, we can show that a primitive strongly regular graph of even order and valency k 9 is 4-extendable. Similarly, we can show that every strongly regular graph of odd order and valency k 3 is 1-nearextendable. Also, there is exactly one primitive strongly regular graph with k 3 which is not 2-near-extendable, namely the Paley graph on 9 vertices (the unique (9, 4, 1, 2)-srg).

The extendability of some specific strongly regular graphs
In this section, we determine the extendability of several families of strongly regular graphs. In the first three subsections, we show that there are many strongly regular graphs with extendability equal or slightly larger than k/2 − 1. In the last subsection, we show that the extendability of any known triangle-free strongly regular graph of even order and valency k equals k − 2.
The reason that the graphs considered in the next three subsections (except for the graphs in Theorem 30) are not k/2 -extendable (when v is even) or not k/2-nearextendable (when v is odd) is the following. Consider the first subconstituent Γ 1 (x) of any fixed vertex x; this is the subgraph induced by N (x). If v is even, we will show that Γ 1 (x) has a matching of size k/2 if k is even and of size (k − 1)/2 if k is odd. When k is odd, there is one vertex y not covered by the matching of size (k − 1)/2 and we choose a vertex z not adjacent to x such that z is adjacent with y. In each case, we construct a matching of size k 2 that cannot be contained in a perfect matching since its removal leaves x isolated. If v is odd, then k is even. We will show that Γ 1 (x) has a matching of size k/2. Choose a vertex y ∈ N 2 (x). The matching of size k/2 in Γ 1 (x) does not cover y. Thus, we construct a matching that cannot be contained in a near perfect matching that misses y since the removal of N (x) ∪ {y} leaves x isolated. We will also use the following lemma.
Lemma 23. Let G be a graph of order am whose vertex set can be partitioned into m subsets, A 1 , A 2 , . . . , A m with equal size a, such that for 1 i m, A i induce a clique, and the graph obtained by vertex contracting each A i contains a perfect matching (when m is even) or a near perfect matching (when m is odd). Then G contains a perfect matching if am is even, and G contains a near perfect matching if am is odd.
Proof. If a is even, the lemma is obvious. If a is odd and m is even, we can find a matching u 1 u 2 , · · · , u m−1 u m such that u i ∈ A i for 1 i m. Now, each subgraph induced by A i \u i contain a perfect matching. Thus G contains a perfect matching. If a is odd and m is odd, we can find a matching u 1 u 2 , · · · , u m−2 u m−1 such that u i ∈ A i for 1 i m − 1. Now, each subgraph induced by A i \ u i contain a perfect matching for 1 i m − 1 and A m contains a near perfect matching. Thus G contains a near perfect matching.

Triangular graphs
. On the other hand, as 2r

Block graphs of Steiner systems
A 2-(n, K, 1)-design or a Steiner K-system is a point-block incidence structure on n points such that each block has K points and any two distinct points are contained in exactly one block. The block graph of such a Steiner system has as vertices the blocks and two distinct blocks are adjacent if they intersect. The block graph of a Steiner K-system is a n(n−1) Theorem 26. Let G be the block graph of a Steiner K-system on n points such that n(n−1) K(K−1) is even. If K ∈ {3, 4} and n > K 2 or K 5 and n > 4K 2 + 5K + 24 + 96 K−4 , the extendability of G is k/2 − 1, where k is the valency of G.
Proof. Let G be the block graph of a Steiner K-system, and let B denote the block sets of the 2-(n, K, 1) design.

. , K}) into cliques, which is
For any A i and A j , there exist b i ∈ A i and b j ∈ A j such that n ∈ b i and n ∈ b j . So, b i and b j are adjacent. The graph obtained by contracting each A i is a complete graph. By Lemma 23, the first subconstituent Γ 1 (x) contains a perfect matching or a near perfect matching. There are k/2 independent edges incident with all N (x) and not incident with x. This implies G is not k/2 -extendable.
Assume that G is not ( k/2 −1)-extendable. By Lemma 3, there is a subset of vertices S such that S contains k/2 − 1 independent edges and r = o(G − S) |S| − 2( k/2 − 1) + 2. Let O 1 , O 2 , . . . , O r be all the odd components of G − S, and P i denote the union of the blocks corresponding to the vertices of O i , where 1 i r. Since |P i | K and P i ∩ P j = ∅ for i = j, we have that n |P 1 | + |P 2 | + . . . + |P r | Kr.
If r 2, then as |S| k by Theorem 5, we get that 2 o(G − S) |S| − 2( k/2 − 1) + 2 k − 2 k/2 + 4 3, contradiction. Otherwise, if r 3 and there exists two singleton components among O 1 , . . . , O r , then |S| 2k − µ. This implies that n/K r contradiction. Note that when K ∈ {3, 4} and n K 2 , the block graph of Steiner K-system is either a complete graph or a complete multipartite graph. If v = n(n−1) K(K−1) is odd, then k = K(n−K) K−1 the electronic journal of combinatorics 21(2) (2014), #P2.34 must be even. The proof of the next result is similar to the proof of Theorem 26 and will be omitted.
Theorem 27. Let G be the block graph of a Steiner K-system on n points such that n(n−1) K(K−1) is odd. If K ∈ {3, 4} and n > K 2 or K 5 and n > 4K 2 + 5K + 24 + 96 K−4 , the near-extendability of G is k/2 − 1, where k is the valency of G.

Latin square graphs
An orthogonal array OA(t, n) with parameters t and n is a t × n 2 matrix with entries from the set [n] = {1, . . . , n} such that the n 2 ordered pairs defined by any two distinct rows of the matrix are all distinct. It is known that an orthogonal array OA(t, n) is equivalent to the existence of t − 2 mutually orthogonal Latin squares. Given an orthogonal array OA(t, n), one can define a graph G as follows. The vertices of G are the n 2 columns of the orthogonal array and two distinct columns are adjacent if they have the same entry in one coordinate position. The graph G is an (n 2 , t(n−1), n−2+(t−1)(t−2), t(t−1))-srg. Any strongly regular graph with such parameters is called a Latin square graph (see [6,Section 9.1.12], [24,Section 10.4] or [37,Chapter 30]). When t = 2 and n = 4, such a graph must be the line graph of K n,n which is also the graph associated with an orthogonal array OA(2, n) (see [7, page 123]) Theorem 28. Let n and t be two integers such that n is even and n 2t 6. If G is a Latin square graph corresponding to an OA(t, n), then the extendability of G is k/2 − 1, where k is the valency of G.
Proof. Let C denote the column set of the orthogonal array OA(t, n) corresponding to G. Consider the neighborhood N (c 1 ) of a column c 1 = (c 1 (1), . . . , c 1 (t)) T of C. There is a partition of N (c 1 ) into cliques, which is A i = {c ∈ C | c(i) = c 1 (i)} for 1 i t. Let l ∈ [n] such that l = c 1 (3). There exist c ∈ A 1 such that c(3) = l and there is c ∈ A 2 such that c (3) = l. Thus c and c are adjacent. The graph obtained by contracting each A i is a complete graph. By Lemma 23 and the same argument in Theorem 26, we deduce that G is not k/2 -extendable.
Theorem 29. Let n and t be two integers such that n is odd and n 2t 6. If G is a Latin square graph corresponding to an OA(t, n), then the near-extendability of G is k/2 − 1, where k is the valency of G.
The line graph of K n,n is a (n 2 , 2(n − 1), n − 2, 2)-srg. It can be regarded as a strongly regular graph corresponding to an OA(2, n).
Theorem 30. Let n be an even integer such that n 4. If G is the line graph of K n,n , the extendability of G is k/2 = n − 1.
Proof. The first subconstituent Γ 1 (x) of some vertex x of G is the disjoint union of two cliques of odd order. Pick two vertices y and z that are not adjacent to x. If S = N (x) ∪ {y, z}, then S contains a matching of size n that is not contained in any perfect matching. Therefore, G is not n-extendable.
Assume that G is not k/2-extendable. By Lemma 3, there is a subset of vertices S such that S contains k/2 independent edges (therefore, S is not the neighborhood of some vertex) and r . , x n } will form an independent set of size n. The set I can be regarded as a perfect matching in K n,n . Any edge in K n,n outside this perfect matching will have non-empty intersection with two edges of this perfect matching. Hence, any vertex in the line graph of K n,n but not in I will be adjacent with two vertices in I. If O 1 is not a singleton, then there is a vertex y ∈ O 1 , where y = x 1 and y is adjacent to x i for some i = 1. This contradicts the fact that O 1 and O i are two distinct components in G − S.
Theorem 31. Let n be an odd integer such that n 3. If G is the line graph of K n,n , the near-extendability of G is k/2 − 1 = n − 2.
Proof. The first subconstituent Γ 1 (x) of some vertex x of G is the disjoint union of two cliques of even order. Thus, Γ 1 (x) contains a matching of size n − 1 and this will imply that G is not (n − 1)-near-extendable. The proof that G is (n − 2)-near-extendable is similar to the proof of Theorem 30 and will be omitted.
Proof. Let G be the (16, 5, 0, 2)-srg. We first show that G is not 4-extendable. Let x be a vertex of G. It is known that the second subconstituent Γ 2 (x) of x is isomorphic to the Petersen graph. Consider four independent edges of Γ 2 (x). We claim that these four edges are not contained in a perfect matching of G. Let S be the complement of the neighborhood of x in G. Then S contains four independent edges and 5 = o(G − S) |S| − 2 · 4 + 2 = 5. Lemma 3 implies that G is not 4-extendable.
We show that G is 3-extendable by contradiction. Assume that G is not 3-extendable. Lemma 3 implies that G has a vertex subset S, such that S contains 3 independent edges, and o(G − S) |S| − 4. Let S be such disconnecting set with maximum size. By the same argument in the proof of Theorem 21, any non-singleton component of G − S cannot contain exactly 3 vertices. If G − S has no singleton components, then G − S has at most two odd components (since α(G) = 5 and each non-singleton component has two non-adjacent vertices). Thus, |S| o(G − S) + 4 6. Lemma 10 implies that |S| 25/2, a contradiction. If G − S has one or two singleton components, then G − S has at most three odd components. Thus, |S| o(G − S) + 4 7. However, S contain the neighborhood of a vertex, which is an independent set of size 5. Since S also contains three independent edges, |S| 8, contradiction. The remaining case is when G − S has at least three singleton components, say x, y, z. As o(G − S) 5, |S| o(G − S) + 4 9. However, |N ({x, y, z})| = 10. This is because y, z are contained in Γ 2 (x) which is isomorphic to the Petersen graph. Because y and z are not adjacent, they have one common neighbor in Γ 2 (x). Hence, there are five vertices adjacent to y or z in Γ 2 (x). Thus, |S| |N ({x, y, z})| = 10, contradiction.
We show that G is 5-extendable by contradiction. Assume that G is not 5-extendable. By Lemma 3, there exists a subset of vertices S such that S contains five independent edges and o(G − S) |S| − 8. Consider such a set S of maximum size. By the same argument as in the proof of Theorem 21, the maximality of |S| implies that any non-singleton odd component of G − S cannot be bipartite. We use this observation to prove that any nonsingleton odd component must has at least 7 vertices. Assume that C is a non-singleton component with 5 vertices ( The (56, 10, 0, 2)-srg is known as the Gewirtz graph or the Sims-Gewirtz graph and its independence number is 16 (see [5, page 372] or [7, page 117]).
Proof. Let G be the Gewirtz graph. We first show that G is not 9-extendable. Let x and y be two non-adjacent vertices of G. Because every vertex in N (x) \ N (y) has exactly 2 neighbors in N (y) \ N (x) and every vertex in N (y) \ N (x) has exactly 2 neighbors in N (x) \ N (y), we can find 8 independent edges with one endpoint in N (x) \ N (y) and the other endpoint in N (y) \ N (x). Let z ∈ N (x) ∩ N (y) and let w be a neighbor of z that is not x nor y. Assume that C is a non-singleton component with 5 vertices. If C has no odd cycles, then C is a bipartite graph and we can add two vertices of C to S. Then |S| will increase by 2 and o(G − S) will increase by 2, contradicting the maximality of S. If Each vertex in the second subconstituent of x has 16 neighbors in N (x) ∩ N (y). By Hall's Marriage Theorem, we can find five independent edges w i u i such that u i ∈ N 2 (x) and w i ∈ N (x) ∩ N (y) for 1 i 5. Let S = N (x) ∪ N (y) ∪ {w 1 , w 2 , w 3 .w 4 , w 5 }. It follows that S contains 21 independent edges, |S| = 43 and o(G − S) 3 = |S| − 21 × 2 + 2. Lemma 3 implies that G is not 21-extendable.
If G were not 20-extendable, by Lemma 3, there is a subset of vertices S such that S contains 20 independent edges and o(G − S) |S| − 38. As before, we may assume that S is such a disconnecting set with maximum size. Then any non-singleton odd component of G − S has at least 5 vertices. Furthermore, we can prove that any non-singleton odd component has at least 7 vertices. Assume that C is a non-singleton component with 5 vertices. If C has no odd cycle, then C is bipartite graph and we can add two vertices from the same color class of C into S. Then |S| will increase by 2 and o(G − S) will increase by 2, contradicting the maximality of |S|. If C induces a pentagon, then Since S contains the neighborhood of a vertex, which is an independent set, in order for S to contain 20 independent edges, S must contain another 20 vertices. Thus, |S| 42, a contradiction.
If G − S only has singleton odd components, then G − S has no even components. Otherwise, we can put one vertex of an even components into S. In this way, |S| will increase by one, and o(G−S) will increase at least by one, contradicting the maximality of |S|. Thus, |S| = 100−o(G−S) 78 which contradicts with |S| o(G−S)+38 60.
We also determine the near-extendability of the known triangle-free strongly regular graphs with odd order. The (5, 2, 0, 1)-srg is precisely 0-near-extendable. The only other known triangle-free strongly regular graph of odd order is the M 22 graph which is (77, 16, 0, 4)-srg with independence number 21 (see [7, page 118]).
Proof. Let G denote the M 22 graph. Let x and y be two non-adjacent vertices. Because every vertex in N (x) \ N (y) has exactly 4 neighbors in N (y) \ N (x) and every vertex in N (y) \ N (x) has exactly 4 neighbors in N (x) \ N (y), we can find 12 independent edges with one endpoint in N (x) \ N (y) and the other endpoint in N (y) \ N (x). Every vertex from N (x) ∩ N (y) has exactly 14 neighbors in the second subconstituent of x. We can find two independent edges w i u i such that u i is in the second subconstituent of x and w i ∈ N (x) ∩ N (y) for 1 i 2. Let S = N (x) ∪ N (y) ∪ {w 1 , w 2 }. It follows that S contains 14 independent edges, |S| = 30 29 and o(G − S) 3 = |S| − 14 × 2 + 1. Lemma 4 implies that G is not 14-near-extendable.
If G were not 13-near-extendable, by Lemma 4, there is a subset of vertices S such that |S| 27 and S contains 13 independent edges and o(G−S) |S|−25. As before, we may assume that S is such a disconnecting set with maximum size. Then any non-singleton odd component of G − S has at least 5 vertices. Furthermore, we can prove that any non-singleton odd component has at least 7 vertices. Assume that C is a non-singleton component with 5 vertices. If C has no odd cycle, then C is bipartite graph and we can add two vertices from the same color class of C into S. Then |S| will increase by 2 and o(G − S) will increase by 2, contradicting the maximality of |S|. If Since S contains the neighborhood of a vertex, which is an independent set, in order for S to contain 13 independent edges, S must contain another 13 vertices. Thus, |S| 29, a contradiction. If t = 0, then o(G − S) = 1. It is also impossible since by our assumption o(G − S) |S| − 25 2.
If G − S only has singleton odd components, then G − S has no even components. Otherwise, we can put one vertex of an even components into S. In this way, |S| will increase by one, and o(G−S) will increase at least by one, contradicting the maximality of |S|. Thus, |S| = 77 − o(G − S) 56 which contradicts with |S| o(G − S) + 25 46.

Final Remarks
The extendability of a strongly regular graph is not determined by its parameters. The Shrikhande graph and the line graph of K 4,4 both have parameter set (16,6,2,2). The extendability of the Shrikhande graph is 2 and the extendability of L(K 4,4 ) is 3. However, we find it remarkable that the extendability of every known primitive triangle-free strongly regular graph of valency k and even order, equals k − 2.
We make the following conjecture regarding the extendability properties of strongly regular graphs of valency k.

Conjecture 37.
If G is a primitive strongly regular graph of valency k, then its extendability (or near-extendability) is at least k/2 − 1.
Note that this conjecture is not true for imprimitive strongly regular graph. For example, the extendability of K a×3 is a/2 = k/4. The conjecture above would be essentially best possible since there are many strongly regular graphs of valency k that are not k/2extendable. If Γ is a (v, k, λ, µ)-srg with λ > θ 2 , then the first subconstituent Γ 1 (x) of any vertex x is connected by eigenvalue interlacing. If G is not a conference graph, then λ − θ 2 1 as θ 2 is an integer. The first subconstituent Γ 1 (x) is λ-regular with second largest eigenvalue at most θ 2 . By [15], Γ 1 (x) contains a matching of size k/2 . If k is even, then this matching cannot be extended to a maximum matching of G. If k is odd, one can add one disjoint edge to this matching such that the result matching of size k 2 cannot be extended to a maximum matching of G. If G is a conference graph with parameters (4t + 1, 2t, t − 1, t)-srg, λ − θ 2 > 1 when t 4. If t = 2 or t = 3, then the first subconstituent contains a matching of size t that cannot be extended to a maximum matching of G. We also remark that there are strongly regular graphs Γ such that the first subconstituent Γ 1 (x) does not contain a matching of size k/2 for any vertex x. For the electronic journal of combinatorics 21(2) (2014), #P2.34 example, if Γ 1 (x) is a disjoint union of cliques K λ+1 and λ is even, then Γ 1 (x) will not contain a matching of size k/2 .
It would be nice to use the extendability properties of strongly regular graphs to study the edge-chromatic number of such graphs of even order. Results from [6,15] imply that any k-regular graph with second largest eigenvalue θ 2 contains at least (k − θ 2 )/2 edge disjoint perfect matchings. It would be interesting to improve this bound for strongly regular graphs.
Counting perfect matchings in regular graphs is an important problem in discrete mathematics (see [22,28]) and a well-known conjecture (see [28,Conjecture 8.18]) states that for any k 3, there exists positive constants c 1 (k) and c 2 (k) such that any kregular 1-extendable graph of order v contains at least c 2 (k)c 1 (k) v perfect matchings (also c 1 (k) → ∞ as k → ∞). Seymour (see [20]) showed that k-regular (k − 1)-edge-connected graphs of order v contains at least 2 (1−1/k)(1−2/k)v/3656 perfect matchings. It would be nice to improve these estimate for strongly regular graphs.
Investigating the extendability properties of distance-regular graphs is also an interesting problem that we leave for a future work. One can deduce from the work of Brouwer and Koolen [8] and Plesník [31] that every distance-regular graph of even order is 1extendable, but it is quite possible that the extendability of many distance-regular graphs is much larger.